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The Easy Peasy Binomial Expansion Trick

October 15th, 2012 by Steven Pomeroy | Filed under Math Tricks.




 

In a previous post, I showed you how to use Pascal’s triangle to quickly expand binomial equations.  There is another math trick that will help you to expand binomials fast.  No – it is not the now infamous joke that many of you have now seen – i.e.:

 

Expand:

(4x + 16y)7

 

Answer:

(4x  +  16y)7

 

(4x   +   16y)7

 

(4x    +    16y)7

 

(4x     +     16y)7

 

(4x      +      16y)7

 

(4x       +       16y)7

 

 

Funny, funny!  But seriously, if you are in need to quickly expand a binomial, and you do not have a detailed Pascal’s triangle readily available, this binomial expansion trick will work out very well for you.

 

First, let’s set up as our example problem:

 

(x – y)5

 

To expand, first start with the x’s.  Begin by writing xN (where N is the exponent the x + y term is being raised to), and continue to xN-1, xN-2 … xN-N:

 

x5               x4               x3               x2               x1               x0

 

Note that x0 is simply equal to 1.

 

Now do the same for the y terms, but in reverse order:

 

x5y0              x4y1              x3y2              x2y3              x1y4              y5

 

Again, note that y0 is equal to 1.

 

The next step is to write in the “+” or “-“ operators.  Since the operation within the parentheses is a subtraction, alternate the operators, beginning with the “+” operator.  Here, the x5y0 is positive, and our example becomes:

 

 

x5     –      x4y1     +      x3y2     –       x2y3     +      x1y4     –      y5

 

Note: if the terms within the parentheses were added (i.e., x + y), then only the “+” operator is used for all terms.

 

Now, write down the term number of each term below each term.  For small exponents, this may not be necessary for you to do, but for large expansions, it is best to do this so that you do not lose track of your term numbers:

 

x5     –      x4y1     +      x3y2     –       x2y3     +      x1y4     –      y5

1               2                 3                   4                  5               6

 

So now comes the part where the coefficients for each term are written.  This is very easy to do with the way we set up our example.

 

Firstly, the coefficients for the first term and the last term are always 1.  For the coefficients of the other terms, simply take the coefficient of the previous term, multiply that by the “x” exponent of the previous term, and then divide the result by the term number of the previous term.  So in our example, the coefficient of the second term would be:

 

(1 x 5)/1 = 5

 

Now write down the coefficient and continue on:

 

x5     –      5x4y1     +      x3y2     –       x2y3     +      x1y4     –      y5

1               2                   3                   4                  5               6

 

 

The coefficient of the third term is:

 

(5 x 4)/2 = 10

 

x5     –      5x4y1     +      10x3y2     –       x2y3     +      x1y4     –      y5

1                2                      3                   4                  5                6

 

 

The coefficient of the fourth term is:

 

(10 x 3)/3 = 10

 

x5     –      5x4y1     +      10x3y2     –       10x2y3     +      x1y4     –      y5

1                 2                      3                       4                  5               6

 

 

The coefficient of the fifth term is:

 

(10 x 2)/4 = 5

 

x5     –      5x4y1     +      10x3y2     –       10x2y3     +      5x1y4     –      y5

1                 2                      3                       4                   5                6

 

Again, the coefficient of the last term is always 1, so our binomial expands out to:

 

x5 – 5x4y1 + 10x3y2 – 10x2y3 + 5x1y4 – y5   answer

 

 

Not so bad, eh?  Good!  Now you can practice this technique a couple of times and you will soon be very efficient at it.  Remember, for more complicated terms, you can use substitution while applying this technique, and then plug back in after you expand the binomial (e.g., if your terms within the parentheses were “4.2w + 33z”,  set x=4.2w and y=33z, and expand it as “x + y”).

 

Below are a couple of binomials together with their expansions.  Try to apply this technique to these binomials and see if your answer agrees with the answers given.

 

 

Expand:

(x + y)4

 

Answer:

x4 + 4x3y + 6x2y2 + 4xy3 + y4

 

 

Expand:

(4x + 16y)7

 

Answer:

(4x)7 + 7(4x)6(16y) + 21(4x)5(16y)2 + 35(4x)4(16y)3

+ 35(4x)3(16y)4 + 21(4x)2(16y)5 +7(4x)(16y)6 + (16y)7

 

 

Expand:

(x – y)9

 

Answer:

x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4 – 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9

 

 

 

 

 

 

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4 Responses to “The Easy Peasy Binomial Expansion Trick”

  1. Dorado | 16/10/12

    GREAT Tool !!!

  2. Jim R. | 16/10/12

    That is impressive.

  3. The Binomial Theorem and Pascal’s Triangle | Math Tricks | 2/11/12

    […] For an alternative method of expanding polynomials, please check out The Easy Peasy Binomial Expansion Trick. […]

  4. VAIBHAV | 14/11/13

    THAT’S GOOD I WANT TO KNOW SOME TRICKS ABOUT DIFFRENTIATION.

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